Force and Potential energy funtion

1. The so-called Yukawa potential
(attached)
gives a fairly accurate description of the interaction between nucleons[that is, neutrons and protons, the constituents of the nucleus]. The constants r0 = 1.5x10-15metres and U0 = 50 MeV.[a] Find the corresponding expression for the force of attraction.

2. Fr= -dU/dr

3. When i finnaly calculate it, the answer is supposed to be negative because
Fr= - dU/dr
= - d (U is given negative)/dr
{ -ve and -ve cancels out }

Fr= +ve d(U)/dr
after integrating U, a -ve sign appears...
so...shouldnt the answer be negative? but the actual solution is a positive. can u tell me y ?

Staff: Mentor

When you find the force by via -dU/dr, the resulting sign will tell you which way the force points (with respect to the variable r). Since it's an attractive force, that will be negative. The magnitude of the force of attraction will be positive, of course.